(a) Prove that a complex differentiable map, $f(z)$, is conformal, i.e. preserves angles, provided a certain condition holds on the first complex derivative of $f(z)$.

(b) Let $D$ be the region

$$D:=\{z \in \mathbb{C}:|z-1|>1 \text { and }|z-2|<2\}$$

Draw the region $D$. It might help to consider the two sets

$$\begin{aligned} &C(1):=\{z \in \mathbb{C}:|z-1|=1\} \\ &C(2):=\{z \in \mathbb{C}:|z-2|=2\} \end{aligned}$$

(c) For the transformations below identify the images of $D$.

Step 1: The first map is $f_{1}(z)=\frac{z-1}{z}$,

Step 2: The second map is the composite $f_{2} f_{1}$ where $f_{2}(z)=\left(z-\frac{1}{2}\right) i$,

Step 3: The third map is the composite $f_{3} f_{2} f_{1}$ where $f_{3}(z)=e^{2 \pi z}$.

(d) Write down the inverse map to the composite $f_{3} f_{2} f_{1}$, explaining any choices of branch.

[The composite $f_{2} f_{1}$ means $f_{2}\left(f_{1}(z)\right)$.]